Question

4 cos^2 x + 4 sin x - 1 = 0

Answer 1

$4 \cdot (1 - \sin^2 x) + 4 \sin x -1=0 \\ \\ 4 -4\sin^2 x +4 \sin x- 1=0 \\ \\ 4\sin^2 x -4\sin x-3=0 \\ \\ t=\sin x \ \ (-1 \leq t \leq 1) \\ \\ 4t^2 -4t-3=0; \ \ \ t_{1,2} = \frac{4 \pm \sqrt{4^2- 4 \cdot 4 \cdot (-3)}}{2 \cdot 4}= \frac{4 \pm \sqrt{16+48}}{8}=\frac{4 \pm 8}{8} \\ \\ t_1=\frac{3}{2} \ \textgreater \ 1; \ \ \ \ \ t_2=-\frac{1}{2} \\ \\ \sin x = -\frac{1}{2} \\ \\ x= -\frac{\pi}{6} + 2 \pi k, \ k \in Z; \ \ \ \ \ \ \ x=\frac{7\pi}{6} + 2 \pi n, \ n \in Z$